Cezar Lupu and Darij Grinberg, Problem
O49, Mathematical Reflections
3/2007. (Link to the solution. For the problem
see 3/2007.)
Proposed solution as a PDF file.
Let A1, B1,
C1 be points on the
sides BC, CA, AB of a triangle ABC. The lines AA1, BB1,
CC1 intersect the
circumcircle of triangle ABC at the points A2, B2,
C2, respectively (apart
from A, B, C). Prove that AA1
/ A1A2
+ BB1 / B1B2 + CC1
/ C1C2
>= 3s2 / (r (4R+r)), where s, r, R are the
semiperimeter, inradius, and circumradius of triangle
ABC, respectively.
Darij Grinberg, The Neuberg-Mineur
circle, Mathematical
Reflections 3/2007.
PDF file. See http://reflections.awesomemath.org/2007_3/NeubergMineur.pdf
for the same file on the Mathematical Reflections server.
In a Mathésis article from 1931,
V. Thébault and A. Mineur established a remarkable
property of quadrilaterals:
Let ABCD be a quadrilateral, and let X, Y, Z, W be the
points on the lines AB, BC, CD, DA which divide the sides
AB, BC, CD, DA externally in the ratios of the squares of
the adjacent sides, i. e. which satisfy
AX / XB = - DA2 / BC2; BY / YC = -
AB2 / CD2; CZ / ZD = - BC2
/ DA2; DW / WA = - CD2 / AB2,
where the segments are directed.
Then, the points X, Y, Z, W lie on one circle.
In the above note, I show a proof of this fact (not
having access to the original paper, I cannot decide
whether this proof is new) as well as an additional
result: If ABCD is a cyclic quadrilateral, then the
circle through the points X, Y, Z, W degenerates into a
line. This additional result is proven in two different
ways, one of them yielding a characterization of this
line as a radical axis.
Thanks to the help of Francisco Bellot Rosado, I have
obtained the Mathésis article
by V. Thébault and A. Mineur referenced as [6] in my
article. It provides an algebraic proof of the
concyclicity of X, Y, Z, W.
Darij Grinberg and Alexei
Myakishev, A
Generalization of the Kiepert Hyperbola,
Forum Geometricorum 4 (2004) pages 253-260.
Consider an arbitrary point P in the plane of triangle
ABC with cevian triangle A1B1C1.
Erecting similar isosceles triangles on the segments BA1, CA1,
CB1, AB1,
AC1, BC1,
we get six apices. If the apices of the two isosceles
triangles with bases BA1
and CA1 are connected by
a line, and the two similar lines for B1
and C1 are drawn, then
these three lines form a new triangle, which is
perspective to triangle ABC. For fixed P and varying base
angle of the isosceles triangles, the perspector draws a
hyperbola. Some properties of this hyperbola are studied
in the paper.
Atul Dixit and Darij Grinberg, Orthopoles
and the Pappus theorem, Forum
Geometricorum 4 (2004) pages 53-59.
If the vertices of a triangle are projected onto a given
line, the perpendiculars from the projections to the
corresponding sidelines of the triangle intersect at one
point, the orthopole of the line with respect to the
triangle. We prove several theorems on orthopoles using
the Pappus theorem, a fundamental result of projective
geometry.
Darij Grinberg, On
the Kosnita Point and the Reflection Triangle,
Forum Geometricorum 3 (2003) pages 105-111.
The Kosnita point of a triangle is the isogonal
conjugate of the nine-point center. We prove a few
results relating the reflections of the vertices of a
triangle in their opposite sides to triangle centers
associated with the Kosnita point.
See also a
geometry-college posting related to the paper.
Darij Grinberg and
Paul Yiu, The Apollonius Circle as a Tucker
Circle,
Forum Geometricorum 2 (2002) pages 175-182.
We give a simple construction of the circular
hull of the excircles of a triangle as a Tucker circle.
See also the
Hyacinthos message related to the paper.
I have also written some other geometric notes
for various purposes (mostly answering questions about how a
theorem is proven). They consist by far not only of new and
original results. The list may grow in the future.
Darij Grinberg, An algebraic approach to
Hall's matching theorem
(version 6 October 2007).
PDF file.
This note is quite a pain to read, mostly due to its
length. If you are really interested in the proof, try the abridged version first.
Hall's matching theorem (also called marriage theorem)
has received a number of different proofs in
combinatorial literature. Here is a proof which appears
to be new. However, due to its length, it is far from
being of any particular interest, except for one idea
applied in it, namely the construction of the matrix S.
See the
corresponding MathLinks topic for details.
It turned out that the idea is not new, having been
discovered by Tutte long ago, rendering the above note
pretty useless.
Darij Grinberg, The Vornicu-Schur
inequality and its variations
(version 13 August 2007).
PDF file. A copy can be found in
a MathLinks article.
This note is dedicated to the so-called Vornicu-Schur
inequality, which states that x(a-b)(a-c) + y(b-c)(b-a) +
z(c-a)(c-b) >= 0 holds whenever a, b, c are reals and
x, y, z are nonnegative reals satisfying some conditions.
What conditions? For instance, (a >= b >= c and x +
z >= y) is a sufficient condition (covering the most
frequently used condition (a >= b >= c and (x >=
y >= z or x <= y <= z))). Another sufficient
condition is that x, y, z are the sidelengths of a
triangle. An even weaker, but still sufficient one is
that x, y, z are the squares of the sidelengths of a
triangle. A yet different sufficient condition is that
ax, by, cz are the sidelengths of a triangle - or, again,
their squares.
These, and more, conditions are discussed, and some
variations and equivalent versions of the Vornicu-Schur
inequality are shown. The note is not primarily focused
on applications, but a few inequalities that can be
proven using the Vornicu-Schur inequality are given as
exercises.
Darij Grinberg, An unexpected
application of the Gergonne-Euler theorem
(version 22 March 2007).
PDF file.
This note solves a problem from the IMO longlist 1976
proposed by Great Britain:
Let ABC and A'B'C' be two triangles on a plane. Let the
lines BC and B'C' intersect at X, and similarly define
two points Y and Z. Let the parallel to BC through A
intersect the parallel to B'C' through A' at X', and
similarly define two points Y' and Z'. Prove that the
lines XX', YY', ZZ' concur at one point.
The proof uses the Gergonne-Euler theorem about the
ratios in which concurrent cevians of a triangle divide
each other. As a sidenote, a simple equivalence between
this Gergonne-Euler theorem and the van Aubel theorem is
established.
Darij Grinberg, On cyclic quadrilaterals
and the butterfly theorem
(version 16 February 2007).
Zipped PDF file.
The note begins with a slight extension of what is
usually referred to as the "butterfly theorem":
Theorem 1. Let k be a circle with center
O, and let A, B, C, D be four points on this circle k.
Let AC and BD intersect at P. Let g be a line through the
point P such that P is the orthogonal projection of the
point O on this line g. Let the line g intersect AB and
CD at X and Z. Then, the point P is the midpoint of the
segment XZ.
We are giving two proofs of this fact. The first one
applies the Pascal theorem to establish a generalization
by Klamkin. The second one combines an affine theorem,
which can be shown using Ceva and Desargues, with some
properties of cyclic quadrilaterals. Finally, an
application of the first proof is given - a triangle
geometry problem from the St. Petersburg Mathematical
Olympiad 2002.
Support page: List of clickable
references.
Darij Grinberg, Two problems on complex
cosines (version 3 February
2007).
PDF file; a copy (possibly outdated) is also downloadable
from the MathLinks forum as an attachment in the topic
"I cant get it".
This note discusses five properties of sequences of
complex numbers x1, x2, ..., xn
satisfying either the equation
x1 = 1/x1
+ x2 = 1/x2
+ x3 = ... = 1/xn-1 + xn
or the equation
x1 = 1/x1
+ x2 = 1/x2
+ x3 = ... = 1/xn-1 + xn
= 1/xn.
Two of these properties have been posted on MathLinks and
seem to be olympiad folklore.
Darij Grinberg, Isogonal conjugation with
respect to a triangle (version
23 September 2006).
Zipped PDF file; a possibly outdated version
(not ZIPped) is also downloadable from the MathLinks
forum as an attachment in the topic
"Isogonal conjugation with respect to a
triangle".
This note gives a detailed introduction into a part of
triangle geometry, and gives new or simplified proofs of
a few recent results.
First, the notions of isogonal lines or isogonals, and of
isogonal conjugates with respect to a triangle are
introduced. Proofs are given for the existence of the
isogonal conjugate, for the fact that the isogonal
conjugate of the circumcircle is the line at infinity,
for the pedal circle theorem and for the relation between
the isogonal conjugate and reflections in sidelines, as
well as for other properties.
Then, the following result of Antreas P. Hatzipolakis and
Paul Yiu is shown:
Let P be a point in the plane of a triangle ABC. The
lines AP, BP, CP intersect the lines BC, CA, AB at the
points A', B', C'. Let Q be the isogonal conjugate of the
point P wrt the triangle ABC. Then, the reflections of
the lines AQ, BQ, CQ in the lines B'C', C'A', A'B' concur
at one point.
The proof of this result given in this note is a
simplification of an argument by Jean-Pierre Ehrmann (see
the
corresponding MathLinks topic).
Then, an observation of José Carlos Chávez Sandoval (again
first published on the MathLinks forum) is proven:
Let P be an Euclidean point in the plane of a triangle
ABC, and let D, E, F be the reflections of this point P
in the perpendicular bisectors of the segments BC, CA,
AB. Denote by AM, BM, CM
the midpoints of the segments BC, CA, AB, and by DM, EM,
FM the midpoints of the
segments EF, FD, DE. Let Q be the isogonal conjugate of
the point P wrt the triangle ABC, and let Q' be the
complement of the point Q wrt the triangle ABC. Then, the
lines AMDM,
BMEM,
CMFM
pass through the point Q'.
Next, the isogonal conjugates of the incenter, the
excenters, the orthocenter, the circumcenter and the
centroid of the triangle are identified, and a short
introduction into the notion of antiparallels is given.
This part is very easy and well-known, but I have not
often seen it written up with proofs, so this is my
attempt at it.
Finally, we show a fact which is not new but seems to be
underused:
Let ABC be a triangle, and let P and Q be two points on
the perpendicular bisector of the segment BC. Then, the
four equations < ABP = < ACQ, < ACP = < ABQ,
< BCP + < BCQ = < BAC and < PBC + < QBC =
< BAC (where all angles are directed angles modulo
180°) are pairwisely equivalent, and they are all
equivalent to the assertion that the points P and Q are
inverse to each other wrt the circumcircle of triangle
ABC. Besides, if this assertion holds, then the lines AP
and AQ are isogonal to each other wrt the angle CAB.
This is used to prove a simple property of the isogonal
conjugates of Kiepert points.
Darij Grinberg,
The Mitten
point as radical center.
Zipped PS file.
The Mitten point of a triangle, defined as the perspector
of the medial and excentral triangles, is shown to be the
radical center of a variable circle triad. In fact, if
the sidelines BC, CA, AB of a triangle ABC are directed
such that the segments BC, CA, AB are positive, and Ab, Ac,
Bc, Ba,
Ca, Cb
are points on the lines CA, AB, AB, BC, BC, CA,
respectively, fulfilling BBa
= CaC = CCb
= AbA = AAc
= BcB, then the pairwise
radical axes of the circles ABcCb, BCaAc, CAbBa are the lines IaMa,
IbMb,
IcMc,
where Ia, Ib, Ic
are the excenters of triangle ABC and Ma,
Mb, Mc
are the midpoints of BC, CA, AB. The radical center of
the three circles is the Mitten point of triangle ABC.
Darij Grinberg, A tour around
Quadrilateral Geometry.
Zipped PDF file.
The tour begins with an arbitrary quadrilateral ABCD. The
external angle bisectors of its angles enclose another
quadrilateral XYZW, which is shown to be inscribed. Its
circumcenter M turns out to be equidistant from the
perpendiculars to the sidelines of ABCD through the
vertices of XYZW.
Next, we pay attention to the case of a circumscribed
quadrilateral ABCD. The incenter O of ABCD happens to
coincide with the intersection of XZ and YW, and the
midpoint of the segment OM' is equidistant from the
perpendicular bisectors of the sides of ABCD. The proof
idea for this result originates from Marcello Tarquini.
Finally, let ABCD be an arbitrary quadrilateral again. We
consider the quadrilateral A'B'C'D' formed by the
pairwise intersections of the perpendicular bisectors of
adjacent sides of ABCD. This quadrilateral A'B'C'D' is
called the PB-quadrilateral of ABCD.
The above implies that the PB-quadrilateral of a
circumscribed quadrilateral is circumscribed, too.
The tour ends with the proof of a converse of this fact:
If ABCD is an arbitrary, but not inscribed,
quadrilateral, and the PB-quadrilateral A'B'C'D' is
circumscribed, then ABCD is circumscribed, too.
An easy-to-prove lemma states that any quadrilateral is
homothetic to the PB-quadrilateral of its
PB-quadrilateral.
Darij Grinberg, Generalization of the
Feuerbach point.
Zipped PDF file.
Using directed angles modulo 180° (crosses), I prove a
suite of theorems about pedal circles and orthopoles of
lines through the circumcenter of a triangle.
Given a triangle ABC, the midpoints A', B', C' of the
sides BC, CA, AB form the medial triangle A'B'C' of
triangle ABC. If U is the circumcenter of ABC and P is
any point different from U, then the reflections x, y, z
of the line PU in the sidelines of the medial triangle
A'B'C' concur at one point L lying on the nine-point
circle of triangle ABC. The reflections X', Y', Z' of L
in B'C', C'A', A'B' are the feet of the perpendiculars
from A, B, C to PU, and the point L itself is the
orthopole of PU with respect to triangle ABC. Let XYZ is
the pedal triangle of P with respect to triangle ABC;
then L lies on the circumcircle of triangle XYZ. If B'C'
and YZ meet at A", and B", C" are defined
cyclically, then L lies on the lines XA", YB",
ZC". (These results cover the first two Fontene
theorems.) Moreover, if HaHbHc
is the orthic triangle of ABC, then the lines
corresponding to PU in the triangles AHbHc, HaBHc, HaHbC pass through L, too. The
orthocenters D, E, F of triangles AYZ, BZX, CXY coincide
with the points corresponding to P in triangles AHbHc,
HaBHc,
HaHbC.
The points X, E, F, Ha
and L lie on one circle. This and more is established in
the first part of the paper.
In the second part, I consider the case when P is the
orthocenter H of triangle ABC. Then, the point L is the
meet of the Euler lines of triangles AHbHc, HaBHc, HaHbC. It is also shown that the
longest of the segments HaL,
HbL, HcL
equals the sum of the two others, solving a problem of
Victor Thebault in the "American Mathematical
Monthly".
In the third part, I regard the case when P is the
incenter O of triangle ABC. The points X, Y, Z are the
points where the incircle of triangle ABC touches the
sides BC, CA, AB. It is shown that the nine-point circle
of ABC touches the incircle, and the point of tangency is
L. This way of establishing the Feuerbach theorem is
presumably new. The point L is known as the Feuerbach
point of triangle ABC. The line OU being called the
diacentral line of triangle ABC, I show that L lies on
the diacentral lines of triangles AHbHc, HaBHc, HaHbC. If Am,
Bm, Cm
are the midpoints of AO, BO, CO, then L lies on the
circles with diameters XAm,
YBm, ZCm
(Michel Garitte) and on the nine-point circles of
triangles BOC, COA, AOB. Moreover, the reflections of the
line OU in the sidelines of triangle XYZ meet at L. The
orthocenters D, E, F of triangles AYZ, BZX, CXY coincide
with the incenters of triangles AHbHc, HaBHc, HaHbC.
Finally, in the last part of the paper, I return to the
original general case with an arbitrary point P and show
that the angle between the nine-point circle and the
pedal circle of P (this is the circumcircle of the pedal
triangle XYZ of P) is 90° - PBC - PCA - PAB. This
theorem is due to V. Ramaswamy Aiyer and generalizes the
Feuerbach theorem.
Some lemmas from my note "Anti-Steiner points with
respect to a triangle" are used in the proofs.
Darij Grinberg, Anti-Steiner points
with respect to a triangle.
Zipped PDF file.
Using directed angles modulo 180° (crosses), we prove
two results of S. N. Collings:
1. The reflections of a line g passing
through the orthocenter H of a triangle ABC in the
sidelines BC, CA, AB concur at a point on the
circumcircle of ABC.
2. This point also lies on the
circumcircles of triangles AYZ, BZX, CXY, where X, Y, Z
are the reflections of an arbitrary point P lying on g in
BC, CA, AB.
This point where the reflections of g in BC, CA, AB
concur is called the Anti-Steiner point of g
with respect to ABC.
Darij Grinberg, New
Proof of the Symmedian Point to be the centroid of its
pedal triangle, and the Converse.
Zipped PS file. See also New Proof
of the Symmedian Point to be the centroid of its pedal
triangle, and the Converse
for a zipped PDF file.
Numerous discussions in the Hyacinthos newsgroup
(beginning with Clark Kimberling's message #1) were
related to the fact that
a) the symmedian point of a triangle is
the centroid of its pedal triangle,
b) and it is the sole point having this
property.
In this note, I present an apparently new proof of part a)
(using the tangential triangle); furthermore, I extend a
standard proof of part a) (given, e. g.,
in Honsberger's book) to a proof of b).
The note is also are avaliable through the Hyacinthos
Files directory, as the file SymmedianPedal.zip (which
contains a Zipped PS file). This note has been announced
in Hyacinthos message #6237.
Darij Grinberg, On the feet of the
incenter on the perpendicular bisectors.
Zipped PS file.
We consider the perpendiculars OX, OY, OZ from the
incenter O of a triangle ABC to the perpendicular
bisectors of its sides BC, CA, AB. It is shown that
triangle XYZ is oppositely similar to triangle ABC, that
the lines AX, BY, CZ pass through the Nagel point of
triangle ABC, and that one of the segments OX, OY, OZ is
equal to the sum of two others.
Darij Grinberg, A Bundeswettbewerb
Mathematik problem and its relation to the Nagel point of
a triangle.
Zipped PS file. See also A
Bundeswettbewerb Mathematik problem and its relation to
the Nagel point of a triangle for a zipped PDF
file.
A problem in the Bundeswettbewerb Mathematik 2003 asked
to prove the following property of triangles:
In a parallelogram ABCD, points M and N are chosen on the
sides AB and BC in a such way that they don't coincide
with a vertex, and that the segments AM and NC have equal
length. Let Q be the intersection of the segments AN and
CM. Then, DQ bisects the angle ADC.
This problem is solved and used to derive the Nagel
theorem (that the incenter of a triangle is the Nagel
point of the medial triangle).
German version: Eine Aufgabe
aus dem Bundeswettbewerb Mathematik und der Nagelsche
Punkt eines Dreiecks at http://www.dynageo.de/discus/messages/5/112.html.
Darij Grinberg, A theorem on
circumscribed quadrilaterals.
Zipped PS file.
Let ABCD be a circumscribed quadrilateral with incenter
O. The perpendicular to AB through A meets BO at M. The
perpendicular to AD through A meets DO at N. Then MN is
perpendicular to AC. I give two proofs for this result,
including an ingenious synthetic proof by Nikolaos
Dergiades. A degenerate case is also discussed.
Darij Grinberg, On the Lemoine
circumcevian triangle.
Zipped PS file.
A new proof is given for the fact that the symmedian
point of a triangle is the symmedian point of its
circumcevian triangle.
Darij Grinberg, New
Insight on the Ninepoint Circle.
The note is on a new proof that the midpoints of the
sides, the feet of the altitudes and the midpoints
between the vertices and the orthocenter of a triangle
lie on one circle.
This a corrected version of the contents of the
file NewNinepoint.zip in the Files directory of the "Hyacinthos" newsgroup.
Support page: Two messages in the
"Hyacinthos" newsgroup related to the note.
German version: Neuer Zugang zum
Feuerbachkreis at http://www.dynageo.de/discus/messages/5/112.html.
Darij Grinberg, Synthetic
proof of Paul Yiu's excircles theorem.
Zipped PS file. See also Synthetic proof
of Paul Yiu's excircles theorem for a zipped PDF file.
In this note, I give a synthetic proof for Paul Yiu's
excircles theorem, which states that, if ABC is a
triangle with orthocenter H, then the triangle whose
sides are the the polars of A, B, C with respect to the
A-excircle, B-excircle, C-excircle of triangle ABC,
respectively, is perspective to triangle ABC, and the
perspector is H.
An old GIF version of the note can be downloaded from the
Files directory of the
"Hyacinthos" newsgroup under YiuSynth.zip (announced in Hyacinthos
message #6176).
